A man on a motorcycle approaches a brick wall as he sounds his horn at a frequency 400 Hz. The sound he hears reflected back from the wall is at a frequency 408 Hz. At what is the speed is the boy riding his motorcycle toward the wall? Assume the speed of sound in air is 340 m/s.



I'm pretty sure the answer is going to be too fast, but let's find out.

This effect of the apparent frequency of a wave (like a sound wave or even a wave of light) depending on the relative motion of the observer and the thing causing the sound is called the Doppler effect. The general formula is given by

fL=(v±vLv±vS)fSf_L = \left(\frac{v\pm v_L}{v\pm v_S}\right)f_S


  • fLf_L is the frequency heard by the listener.
  • fSf_S is the frequency emitted by the source.
  • vv is the speed of propogation in the medium. In this case that's the speed of sound in air (given to us as 340 m/s).
  • ±vL\pm v_L is the relative velocity of the listener. This is positive when the listener is moving toward the source.
  • ±vS\pm v_S is the relative velocity of the source of the wave. This is positive when the source is moving away from the listener.

Probably the toughest thing about doing the Dopper effect calculations is remembering those sign conventions. You'll probably want to practice those.

Anyway, in this case, we really have to do two calculations. First the boy emits sound and it travels to the wall who is effectively the "listener". Then it reflects the sound, becoming the emitter while the boy becomes the "listener".

Wave travelling toward the wall part

In this case, we have

  • fL=fwallf_L = f_\text{wall}
  • fS=400f_S = 400 Hz
  • v=340v = 340 m/s
  • vL=0v_L = 0. We don't even need a sign here because that wall ain't movin'.
  • vS=vboyv_S = v_\text{boy}. We don't know the boy's speed yet, but it's definitely a negative since he's moving toward the wall.

Putting this together, we have

fwall=400(340340vboy)f_\text{wall} = 400\left(\frac{340}{340- v_\text{boy}}\right)

Wave travelling away from the wall part

In this case, we have

  • fL=408f_L = 408 Hz
  • fS=fwallf_S = f_\text{wall}
  • v=340v = 340 m/s
  • vL=vboyv_L = v_\text{boy}. This time the sign is positive according to our rules. Confusing, right?
  • vS=0v_S = 0. That wall is still just sitting there.

Putting this together, we have

408=fwall(340+vboy340)408 = f_\text{wall}\left(\frac{340+v_\text{boy}}{340}\right)


So now we have a system of two equations in two unknowns. Use whatever methods you prefer for that. You might try eliminating the fwallf_\text{wall} since you don't need that and then solving for vboyv_\text{boy}, but it's up to you. It's just algebra from here. So I'll let you take it from here. It tuns out the boy isn't driving that fast at the wall. But still, you shouldn't drive your motorcycle straight at walls. 🙄

Let me know if you are unclear on any of the steps. Good luck!

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